The base of a triangle is decreasing at a rate of $13$ millimeters per minute and the height of the triangle is increasing at a rate of $6$ millimeters per minute. At a certain instant, the base is $5$ millimeters and the height is $1$ millimeter. What is the rate of change of the area of the triangle at that instant (in square millimeters per minute)? Choose 1 answer: Choose 1 answer: (Choice A) A $8.5$ (Choice B) B $21.5$ (Choice C) C $-8.5$ (Choice D) D $-21.5$
Setting up the math Let... $b(t)$ denote the base of the triangle at time $t$, $h(t)$ denote the height of the triangle at time $t$, and $A(t)$ denote the area of the triangle at time $t$. We are given that $b'(t)=-13$ and $h'(t)=6$ (notice that $b'$ is negative). We are also given that that $b(t_0)=5$ and $h(t_0)=1$ for a specific time $t_0$. We want to find $A'(t_0)$. Relating the measures The measures relate to each other through the formula for the area of a triangle: $A(t)=\dfrac{b(t)h(t)}{2}$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=\dfrac{b'(t)h(t)+b(t)h'(t)}{2}$ Using the information to solve Let's plug ${b'(t_0)}={-13}$, ${h(t_0)}={1}$, ${b(t_0)}={5}$, and $C{h'(t_0)}=C{6}$ into the expression for $A'(t_0)$ : $\begin{aligned} A'(t_0)&=\dfrac{{b'(t_0)}{h(t_0)}+{b(t_0)}C{h'(t_0)}}{2} \\\\ &=\dfrac{({-13})({1})+({5})(C{6})}{2} \\\\ &=8.5 \end{aligned}$ In conclusion, the rate of change of the area of the triangle at that instant is $8.5$ square millimeters per minute. Since the rate of change is positive, we know that the area is increasing.